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anonymous
 3 years ago
Prove that if n is a perfect square, then n +2 is not a perfect square
anonymous
 3 years ago
Prove that if n is a perfect square, then n +2 is not a perfect square

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i suppose first step would be to assume n is perfect square

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then i suppose i find n^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0n + 2 would be m^2 + 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so now... i assume m^2 + 2 is a perfect square...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we have no 2 perfect square numbers with difference 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i let m^2 + 2 = k^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mukushla that's not really a proof.....

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0you may continue @igba

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm i suppose the next step would be m^2  k^2 = 2 so (m+k)(mk) = 2 i don't think this proves anything...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually that is :) \[m^2k^2=2\]\[(mk)(m+k)=2\]and this is impossible

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0well,,you've almost proved it.. now m and k both are integers/.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0and its k^2 m^2 = 2 and not otherwise..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now...integers huh...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[m+k>mk\]so\[m+k=2\]so\[m=k=1\]but it gives\[mk=0\]

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0you know k>m.. so from (k+m)(km)= 1*2 =>k+m =2 and km = 1 this doesnt have an integral solution..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why do i know k > m again?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0m^2 + 2 = k^2 both m and k are positive integers hence k>m

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0in layman language,, k^2  m^2 >0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when was it assumed that m and k are positive?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0what else do you mean by perfect square ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i don't see how that relates.... the definition of a perfect square is x = k^2

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.04 is a perfect sq since its 2^2 i.e. square of a positive integer.. and likewise..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mukushla "we have no 2 perfect square numbers with difference 2" Because 0,1,4,9,6,5,6,9,4,1 etc....

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0you take the absolute value

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0i mean you have to..

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0see,, (k+m)(km) =2 either both are positive are both are negative,,since RHS is postive.. so even if you take ve values,, negative * negative = positive ,,i.e. treated as absolute value..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0...i don't think this is a valid proof....

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0another explanation might me x is a perfect square because its y^2 x = y^2 = (y)^2 sqrt(x) =  y  =  y 

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0who can writing here now two perfect squares with difference of 2 ? so because i think do not exist or .... ???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since when did sqrt x become y

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.00 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats. No +2 in there.

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0so and i think that this difference of two will be the ,,key" of this proof of ...

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0@lgbasallote what is your opinion from this ,please ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my opinion is that this is some tricky algebra

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0how do you think it ???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because i can't think of a proof

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how can you prove/disprove that the difference of two squares is 2...

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0so i think this is very easy logicaly but to prove it will be difficile i think so but there are again @saifoo.khan and @satellite73 probably they will can doing it hope so much

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0their fields of specialty aren't exactly in proving

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What's wrong with my proof?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anyway... m^2 + 2 = k^2 k^2  m^2 = 2 for it to be equal to 2...k^2 and m^2 should either be both even...or both odd so if i assume both are even.. then k and m are also both even so i can rewrite this as (2x)^2  (2y^2) = 2 4x^2  4y^2 = 2 then... 2x^2  2y^2 = 1 x^2  y^2 = 1/2 then since it's x^2 and y^2 are both integers...the difference should also be integer \('\therefore\) contradiction

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So, that doesn't mean it's invalid..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it does actually.....you can't do infinite substitutions to prove anything....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There is no infinity involved, all integer squares end in the numbers I gave.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0infinite loops involves infinity agree?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No infinite loop either, just logic.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0An integer ends in one of digits 0 to 9. The squares must end in etc....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but we both know that setting values is illegal in proving

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am not setting any values.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you're checking the squares of each number

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and though there is a pattern...there is still susbtitution involved

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and then you'll have to substitute forever to verify that the pattern really does not break

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then infinity is involved

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's not a pattern, it's a fact 0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you do admit that you checked the 0 to 9 squares

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Anything else is impossible.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0...that's substitution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anyway...my point is...substitution isn't really allowed in proving

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You will find that this fact is used a lot in many valid proofs in number theory.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am not using brute force, brute force is something like testing or experimental math. Or exhaustion (which is also a valid proof technique). It's the use of a fact about squares (ALL squares)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you prove it? that it's a fact about squares?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, all numbers end in 0 to 9 and so their squares end in .....QED.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you didn't really demonstrate anything...you just repeated what you said

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You asked for proof, I just gave u a proof.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you deny my proof, provide a counterexample.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since you try so hard to defend your proof, i'll accept it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's very gracious, thank you:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do check on my algebra proof though please. i would like to know if that's the way contradiction works

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OK, I will look at it now....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Did u do the case when both are odd? (Perhaps a more straightforward way is one you start with n= some k squared then next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0again i ask... how is k > m

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0factoring involves lesser steps...its obvious that k>m \[k^2−m^2=2\]\[(k−m)(k+m)=2\]easily u can see \[km=1\]\[k+m=2\]add them\[2k=3\] Contradiction

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0again i ask for the nth time @mukushla how is k > m

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok man i just want to answer ur question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"2 can only be an outcome of even and even or odd and odd" is a valid statement Why do you think it might not be valid?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well because 8  4 is not 2 but it's even minus even

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OK, so finally, the algebra Given n=k^2 > (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i thought they are positive :) sorry

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why (k+1)^2 @estudier ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I already, said, that's the next square....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is this still related to my proof?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0? n is k^2, what is the next square?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no problem ... this is better\[k^2m^2=2\]\[(km)(k+m)=2\]so\[km=1\]\[k+m=2\]because this time im sure that \(k>m\) ...now add them\[2k=3\]Contradiction and we are done

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, this proof Given n=k^2 > (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you allowed to do that? @mukushla

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why not...for any real number \[a^2=a^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Given n=k^2 > (k+1)^2 = k^2 + 2k +1 k^2 + 2k +1 >= n +3 n +3 > n+2 QED

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's 2 proofs I have given now...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a few questions in mind @estudier 1) why find the next square? 2) how is k^2 + 2k + 1 > = n + 3 3) what does n + 3 > n+2 prove?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Think about it....:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in the meanwhile... @mukushla why k  m = 1?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because\[km<k+m\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how does that make k  m = 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u have 2 positive integers with ab=2 and a<b what are a and b....clearly 1 and 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0who said those are positive?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0something feels wrong...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0anyway...where did you get 2k > 3

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok...think on what i said...because i believe thats the shortest way to prove ur statement

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0adding\[km=1\]and\[k+m=2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Going back to your question about 2, u can just adjust it to difference of 2q...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im really skeptical about this... especially that k^2  m^2 = 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You were worried about the validity of your statement, change it to 2q instead of just 2.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Or 2p, or whatever letter u like best.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01) why find the next square? Because we want to see that it isn't n+2 2) how is k^2 + 2k + 1 > = n + 3 Substitute n=k^2 3) what does n + 3 > n+2 prove? See 1)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh. when i said difference of two numbers equal to 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and @mukushla how did you get k > m

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, because the next square is (k+1)^2 (which might or might not be equal to n+2, that is what we have to discover.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and since the next square is greater than n + 2...that makes n + 2 invalid for any succeeding squares?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, the algebra is completely general...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i'm still confused about k^2 + 2k + 1 though...if n = k^2...it just makes n + 2k + 1...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0...which is greater than n+3, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Number theorists don't pay attention to 0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Tradition, convention, call it as u like..... So in fact there are 2 definitions for natural number , one with 0 and one without.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0..you give so many weird proofs......the sad part is that i cannot disprove it because i know nothing about it....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"because i know nothing about it" But you do now:)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's not like it's limited to the 99s....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i was referring to your claims earlier like brute force is valid proof in number theory and that number theorists don't consider 0...i cannot disprove your proofs because i do not know if these claims are true or not

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Number theory is less onerous than real number stuff because it is integer only. But it is actually this difference that causes all the difficulty, students are not accustomed to thinking in a more freewheeling way.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let n be a perfect square where n=m^2 then next consecutive perfect square number would be (m+1)^2 Now, (m+1)^2m^2=2m+1 Since, 2m+1>2 as m is a set of natural numbers. Thus, if n is a perfect square..... n+2 cannot be a perfect square as the difference between two perfect squares is always more than 2.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, that's similar to the one I gave...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"number theorists don't consider 0" http://en.wikipedia.org/wiki/Natural_number

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://mathforum.org/library/drmath/view/63510.html Using 0,1,4, etc...

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0ok,so who can answering this question ? ,,the line of perfect squares begin with number ... " what is this number ?

jhonyy9
 3 years ago
Best ResponseYou've already chosen the best response.0so and for this your question @lgbasallote i think that for n=m^2 and n+2 for being a perfect square so than n+2=m^2 +2 suppose m^2 +2 =x^2 like a perfect square so than m^2 = x^2 2 so m1,2 = +/ sqrt(x^2 2)
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